# 6.5: The temperature (heat) equation

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We now resume our quest for a closed set of equations to describe the flow of a Newtonian fluid. We previously assumed mass and momentum conservation, resulting in the density equation and the Navier-Stokes momentum equation Equation 6.3.37. This collection of equations totals 4, but involves 5 unknowns: \(ρ\), \(p\), and the three components of \(\vec{u}\).

We have now invoked a new assumption, namely energy conservation in the form of the 1st law of thermodynamics. This will allow us to add a new equation to the set. We first convert Equation 6.4.29 to Eulerian form. Two terms require conversion to volume integrals.

• We apply Cauchy’s lemma to the left -hand side, resulting in:

\[\frac{D}{D t} \int_{V_{m}} \rho \mathscr{I} d V=\int_{V_{m}} \rho \frac{D \mathscr{I}}{D t} d V.\]

• The heat loss can be converted using the divergence theorem (section 4.2.3):

\[\oint_{A_{m}} \vec{q} \cdot \hat{n} d A=\int_{V_{m}} \vec{\nabla} \cdot \vec{q} d V\]

We now have

\[\int_{V_{m}}\left(\rho \frac{D \mathscr{I}}{D t}+\vec{\nabla} \cdot \vec{q}+p \vec{\nabla} \cdot \vec{u}-\rho \varepsilon\right) d V=0.\qquad .\]

We conclude as usual that the integrand must be zero everywhere, resulting in:

\[\rho \frac{D \mathscr{I}}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}-p \vec{\nabla} \cdot \vec{u}+\rho \varepsilon,\label{eqn:1}\]

where Equation 6.4.27 has been used for the heat flux. We have gained a new equation, but have also introduced two new unknowns, the internal energy I and the temperature \(T\). (Note that neither \(\varepsilon\) nor \(\vec{q}_{rad}\) counts as an unknown. The former is determined by the velocity field via the strain, \(\varepsilon = 2v e^2_{ij}\), while we assume that the latter is specified independently.) This leaves us short two pieces of information.

## 6.5.1 Specific heat capacity

Our next goal is to eliminate internal energy from the problem by establishing a relationship between it and temperature. We will consider two possible approaches, each based on an assumption about the nature of the temperature changes that can be illustrated with a simple lab experiment.

#### Incompressible heating

Suppose that a fluid sample is contained in a closed, rigid vessel wherein it is slowly heated. We keep track of the heat input and the resulting temperature rise, and the two turn out to be approximately proportional:

\[\left(\frac{\partial Q}{\partial T}\right)_{\upsilon}=C_{\upsilon}.\label{eqn:2}\]

Here \(Q\) is the specific heat content, i.e., heat content per unit mass, typically measured in joules per kilogram. The subscript \(\upsilon\) on the partial derivative specifies that the specific volume \(\upsilon=\rho-1\) is held fixed while \(Q\) and \(T\) are changing. \(C_\upsilon\) is called the specific heat capacity at constant volume, and can be regarded as constant if the range of temperatures is not too wide. Typical values are

\[c_{\upsilon}=\left\{\begin{array}{l}

42000 k_{k}-1_{k}-1, \text { in\ water } \\

7000 k g^{-1} K^{-1}, \text {in\ air. }

\end{array}\right.\label{eqn:3}\]

Now because this incompressible fluid does not expand or contract, changes in internal energy are due entirely to changes in heat content, so

\[\frac{D \mathscr{I}}{D t}=\frac{D Q}{D t}=C_{\upsilon} \frac{D T}{D t}.\]

We can now rewrite Equation \(\ref{eqn:1}\) as

\[\rho C_{v} \frac{D T}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}+\rho \varepsilon,\label{eqn:4}\]

keeping in mind that \(\vec{\nabla}\cdot\vec{u}=0\). We have now succeeded in eliminating \(E\), but the solution only works if \(\vec{\nabla}\cdot\vec{u}=0\). Compressibility effects are not always negligible, especially in gases. To allow for that possibility, we imagine a slightly different experiment.

Isobaric heating

Suppose that we once again apply heat to a sample of fluid, but that the fluid is enclosed not in a rigid container but rather in a flexible membrane, like a balloon. As a result, the fluid can expand or contract freely, but the pressure does not change. (We assume that our balloon does not change altitude significantly, as would a weather balloon.) The process is therefore designated as *isobaric*.

For this process we define a new thermodynamic variable called the specific *enthalpy*, \(H\). When a system changes slowly, the change in enthalpy is given by ∆H = ∆I +∆(pυ). In an isobaric process, this becomes \(\Delta H = \Delta \mathscr{I} + p\Delta \upsilon\). For a given change in temperature, the change in enthalpy is given by

\[\left(\frac{\partial H}{\partial T}\right)_{p}=C_{p}.\label{eqn:5\]

\(C_p\) is called the specific heat capacity at constant pressure, and is approximately constant for small temperature changes. Typical values are

\[C_{p}=\left\{\begin{array}{l}

4200 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1}, \text {in water } \\

1000 \mathrm{J} \mathrm{kg}^{-1} \mathrm{K}^{-1}, \text {in air. }

\end{array}\right.\label{eqn:6}\]

For a fluid parcel undergoing this isobaric change, the material derivative of the enthalpy is

\[\frac{D H}{D t}=\frac{D \mathscr{I}}{D t}+p \frac{D \upsilon}{D t}=C_{p} \frac{D T}{D t}.\label{eqn:7}\]

The material derivative of \(\upsilon\) can be written as

\[\frac{D \upsilon}{D t}=\frac{D \rho^{-1}}{D t}=-\rho^{-2} \frac{D \rho}{D t}=-\rho^{-2}(-\rho \vec{\nabla} \cdot \vec{u})=\rho^{-1} \vec{\nabla} \cdot \vec{u},\label{eqn:8}\]

where mass conservation Equation 6.2.5 has been invoked. Multiplying Equation \(\ref{eqn:7}\) by density and substituting Equation \(\ref{eqn:8}\), we have

\[\rho \frac{D H}{D t}=\rho \frac{D \mathscr{I}}{D t}+p \vec{\nabla} \cdot \vec{u}=\rho C_{p} \frac{D T}{D t}\]

which, together with Equation \(\ref{eqn:1}\), gives

\[\rho C_{p} \frac{D T}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}+\rho \varepsilon.\label{eqn:9}\]

## 6.5.2 The heat equation

The two temperature equations that hold in the incompressible and isobaric approximations, Equation \(\ref{eqn:4}\) and \(\ref{eqn:9}\), differ only in the choice of the specific heat capacity:

\[\rho\left(C_{\upsilon} \text { or } C_{p}\right) \frac{D T}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}+\rho \varepsilon.\]

Which approximation is better? In water, there is virtually no difference, because \(C_\upsilon\) is nearly equal to \(C_p\). In air, compressibility can be important, so the isobaric approximation is preferable. For geophysical applications, then, we choose the isobaric version:

\[\rho C_{p} \frac{D T}{D t}=k \nabla^{2} T-\vec{\nabla} \cdot \vec{q}_{r a d}+\rho \varepsilon.\label{eqn:10}\]

This is a generalization of the “heat equation’’ often discussed in physics texts: neglecting the radiation and dissipation terms, it becomes

\[\frac{D T}{D t}=\kappa_{T} \nabla^{2} T.\label{eqn:11}\]

Here, \(\kappa_T\) is the thermal diffusivity, given by

\[\kappa_{T}=\frac{k}{\rho C_{p}}=\left\{\begin{array}{ll}

1.4 \times 10^{-7} m^{2} s^{-1}, & \text {in water } \\

1.9 \times 10^{-5} m^{2} s^{-1}, & \text {in air }

\end{array}\right.\label{eqn:12}\]

By using Equation \(\ref{eqn:10}\) instead of \(\ref{eqn:1}\) we are able to impose energy conservation while adding only one new unknown (instead of two), so at least we are no worse off in terms of closure. We now have 5 equations for 6 unknowns.

In the special case of incompressible fluid, the condition \(\vec{\nabla}\cdot\vec{u}=0\) represents an additional equation and the set is closed (i.e., no more equations are needed). For the more general case, we need to make a further assumption about the nature of the fluid. This will take the form of an equation of state. Again, there is more than one possibility.